Calculus
posted by Reggie .
A trough is triangular in cross section, an isosceles triangle with sides of 12 inches, and a top of 10 inches. The trough is 40 inches long. How fast is the depth changing if you are pumping one cubic foot per minute into the trough?
I am having a hard time getting it in terms of h.

Make a sketch of the crosssection of the isosceles triangle. Draw it some water level, letting the height be h and letting the width of the water level be x
by similar triangles
x/h = 10/12
12x = 10h
x = 5h/6
volume of water = 40(area of triangle)
= 40(1/2)(x)h
= 20(5h/6)(h) = 50/3 h^3
V = (50/3) h^3
dV/dt = 50 h^2 dh/dt
have to watch our units, 1 cubic foot = 12^3 or 1728 cubic inches
1728 = 50 h^2 dh/dt
dh/dt = 1728/(50h^2) = 864/25 inches/minute 
wow I was headed that way didn't watch units Thank You very much
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